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# Finding fluid speed exiting hole

## Video transcript

so where we left off we had this this canister or because it had a closed top and a vacuum above the fluid and the fluid on top had a hole I'm using a big the big marker right now with a fluid on top in an area of a1 and I poked a little hole with it with a super small area a2 and I said the area of a - it's so small it's one one thousandth of area one and then we used the the continuity equation we said well the the velocity the rate at which the surface is moving up here that's v1 times area one right times the whole surface area of the liquid has to be equal to the output velocity which we're trying to figure out as a function of everything else times this output area and then we say well I don't remember let's see I made a mistake so I'm I don't know if I did this in the last video or I did this in my mistake video but anyway so we know that the V the initial the this top velocity times this top area is equal to the output velocity times instead of writing area to we could write area one over a thousand right we could write area one over a thousand and we would still have the area one here and then I don't I you can get rid of the area one on both sides and then you're saying that the output velocity is equal to essentially or sorry the velocity up here is equal to the rate at which the top of the the surface moves down is equal to one one thousandth of the velocity of the liquid spurting out of this little hole so now with that we actually have the three variables for kind of the right hand side or sorry the left hand side of Bernoulli's equation and what are the variables on the left hand side well what is the pressure at this point where we have a hole well and this is an important thing when we talk about you know Bernoulli's let me rewrite Bernoulli's equation it's p1 plus rho rho g h1 plus ro v1 squared over two is equal to pressure two plus Rho G H two plus Rho v2 squared over two we figured out all of these terms so now let's figure out the things that we have to input here so what is the pressure at point two and this is the important thing you might want to say and this was my initial reaction too and that's why I made a mistake is that oh well what's the pressure at this depth in the fluid and that's not what Bernoulli's equation is telling us the Bernoulli's equation is telling us actually what is the external pressure at that hole right what how much how much it when we did the derivation we were saying how much work this was kind of the work term although we played around with a little bit right but if we look at at the water that's burning out of the hole it's not doing any work because it it's not it's not actually exerting force against anything so it's not actually doing work so when we think about the pressure the output pressure it's not the pressure at that depth of the fluid you should think of it as the pressure at the hole the external pressure at the hole and in this case there is no external pressure at the hole if there was let's say let's say that if we if we close the hole then at that point sure the pressure would be the pressure that's being exerted by the outside of the canister to contain the water in which case we would end up with no velocity the water wouldn't sprint anywhere but now we're saying the external pressure zero that's what the hole essentially creates so we're going to say that p2 is zero so this is zero so this pressure was zero because we're in a vacuum and then this and then P 2 is also zero so though both of these are zero remember that's kind of the external pressure it's the pressure P 1 is the pressure external pressure to the input to the pipe and you can kind of view this as a pipe right I could redraw that that Cup is kind of like this well I could redraw it as a pipe like that looks like well I actually know the let me undo that that has a big hole on the top and it goes down some level to a super small hole like that and so and this would be a vacuum and that the fluid is just going in and spurting out at this end right so anyway the pressure going into the pipe is zero and we said since we put a hole the pressure coming out of the pipe is zero so we're doing no work and what is this term well this was this was the potential energy term and we said that H 1 is equal to H is equal to we're saying that this is zero height so now this simplifies to Rho times gravity times h plus Rho times V 1 squared well V 1 we said is equal to this so Rho over 2 times V 2 over a thousand squared all right I just substituted V 2 over a thousand for V 1 that equals the pressure at the hole the external pressure at the hole which is zero plus what is H 2 well this is H 2 right here which we said is zero we determined that the hole was poked at height zero so this is also zero so that equals essentially this kinetic energy like term it's not exactly kinetic energy its Rho times V squared - B squared divided by 2 now one thing that we can immediately see is that we have all these rows on both sides equations so we can divide both sides by Rho we get rid of all of those and then we can multiply both sides of the equation by 2 and we get 2 G H is equal to sorry Plus this term V squared V 2 squared over what's a thousand squared over 1 million over 1 million and that is equal to V 2 squared now we could do a you know we could do the exact thing we could subtract one over a million v2 squared from both sides and we would get you know point nine nine nine nine nine nine nine v2 squared but let's just say for the sake of simplicity or let's say if you know if this wasn't a thousand let's say this was a million so we you know this the surface area is a much bigger we see that this term becomes very very very small and if if that whole is one millionth of the surface area then it becomes really insignificant so we can ignore this this this term at because it just makes things complicated and we're assuming this is a really really large number that this whole is much smaller than that then then the surface area of the fluid and we you know this is like poking a hole in Hoover Dam right Hoover Dam is backing up this huge lake and you poke a hole in it so that hole is going to be this very small fraction of the surface area of the fluid and you can only make this assumption when it is when that output hole is much smaller than the input hole but with that said what is the output velocity well the velocity is going to just take the square root of both sides this is the square root of sorry - gh so that is the output velocity and what is the amount of what is the amount of liquid that flows out per second well we figure that out already it's it's you know kind of like a column of fluid that comes out so per second the length of the column of fluid will be the velocity times time and then the the cross section of that column is equal to the output area so if I wanted to know the flow coming out it would be the flow coming out or the flux coming out would be equal to the out the holes area times the holes the output velocity and that would equal the area times the square root of 2gh and we could use that to actually solve problems in the future yeah but if we had actual numbers anyway I only have a minute and a half left so I'll see you in the next video